Earth and Planetary Materials Spring 2013 Lecture 4 2013.01.16
Example Beryl Be 3 Al 2 (SiO 3 ) 6 Goshenite Aquamarine Emerald Heliodor Red beryl Morganite pure Fe 2+ & Fe 3+ Cr 3+ Fe 3+ Mn 3+ Mn 2+
Rules for ionic (atomic) substitution Size and charge matter!
Victor M. Goldschmidt 1888 1947 Swiss born Norwegian mineralogist and petrologist who laid the foundation of inorganic crystal chemistry and founded modern geochemistry
Goldschmidt s Rules Size Atomic substitution is controlled by size (radii) of the ion If size difference is < 15%: free substitution 15 ~30%: limited substitution > 30%: little to no substitution If there is a small difference of ionic radius, the smaller ion enters, the crystal preferentially
Goldschmidt s Rules Charge Atomic substitution is controlled by charge of the ions If charge difference is: = 1: substitute readily if electrical neutrality is maintained >1 : substitution is minimal For ions of similar radius but different charges, the ion with the higher charge enters the crystal preferentially Forms a stronger bond with the anions surrounding the site
Ringwood s Modifications (1955) Substitutions may be limited, even when the size and charge criteria are satisfied, when the competing ions have different electronegativities and form bonds of different ionic character Explains discrepancies i with respect to the Goldschmidt rules Example: Na + and Cu + have the same radius and charge, but do not substitute for one another
Other factors affecting substitution Temperature Minerals expand at higher T Greater tolerance for ionic substitution at higher T Pressure Increasing pressure causes compression Less tolerance for ionic substitution at higher P Availability of ions ions must be readily available for substitution to occur
Types of substitution Simple substitution Interstitial substitution Vacancy substitution
(1) Simple substitution Involves the same sites in the crystal structure Type 1: Complete binary solid solution: ions have same charge and nearly same size, substitute freely Example: Fe 2+ Mg 2+ Ni 2+ (0.86 Å) (0.80 Å) (0.77 Å)
Example Olivine group End members are forsterite (Mg 2 SiO 4 ; Fo ) and fayalite (Fe 2 SiO 4 ; Fa ) Any composition in between is possible: (Mg x Fe 1 x ) 2 SiO 4 x =0.7 would be 70 mole percent forsterite ; or Fo 70
(1) Simple substitution Type 2: Coupled substitution: ions have nearly same size, but different charge; balance charge by simultaneous substitution on a second site.
Example Si 4+ and Al 3+ (CN = 4): balanced by also substituting Ca 2+ for Na + on a different site write as a charge balanced reaction: Si 4+ + Na + = Al 3+ + Ca 2+ Example Plagioclase feldspars: Albite (NaAlSi 3 O 8 ) Anorthite (CaAl 2 Si 2 O 8 )
Giving isomorphic derivative structures
(2) Interstitial substitution Coupled substitution where the charge balancing substitution is accommodated on a site normally vacant in the mineral Many minerals have large channels or space between layers that can accommodate cations
Example Beryl Be 3 Al 2 (SiO 3 ) 6 Channels in beryl accommodating charged ions
(3) Vacancy (Omission) substitution Charge is balanced by leaving sites vacant Example: pyrrhotite: Fe (1 x) S contains both Fe 2+ and Fe 3+ For charge balance 3(Fe 2+ ) = 2(Fe 3+ ) + where represents and Fe site that is vacant. Another way to represent the formula is: (Fe 2+ ) 1 3x (Fe 3+ ) 2x x x S
Point defects
Point defects under microscope
Summary on substitution
Summary on substitution Simple substitution Complete binary substitution Coupled substitution Interstitial substitution Vacancy substitution Varied composition
Chemical composition
Isomorphs and derivative structures Isomorphs: minerals that have same crystal structure but different chemical composition Example: berlinite is AlPO 4 with the quartz structure (substitute t Si 4+ + Si 4+ = Al 3+ + P 5+ )
Isomorphs Examples KMg 3 (AlSi 3 O 10 )(OH) 2 phlogopite K(Li,Al) 2 3 (AlSi 3O 10)( )(OH)) 2 lepidolite KAl 2 (AlSi 3 O 10 )(OH) 2 muscovite Amphiboles: Ca 2 Mg 5 Si 8 O 22 (OH) 2 tremolite Ca 2 (Mg,Fe) 5 Si 8 O 22 (OH) 2 actinolite Actinolite series minerals (K,Na) 0 1 (Ca,Na,Fe,Mg) 2 (Mg,Fe,Al) 5 (Si,Al) 8 O 22 (OH) 2 Hornblende
From chemical composition to formula Or vice versa
Reporting a mineral/rock composition Element weight %: e.g. 5 g Fe in 100 g sample Oxide weight %: Majority of minerals contain large quantities of oxygen, e.g. silicates, oxides, carbonates Number of atoms mineral chemical formula Relating all elements to one: normalization
From element wt% to formula Example A wet chemical analysis: Na = 32.8 % Al = 12.8 % F = 54.4 % What is the unknown mineral? wt Atomic weight wt / atomic weight mole normalization mole ratio Na 32.8 22.99 32.8/22.99 1.4267 1.4267/0.4744 3.007 ~3 Al 12.8 26.98 12.8/26.98 0.4744 0.4744/0.4744 1.000 ~1 F 54.4 19.00 54.4/19.00 2.8631 2.8631/0.4744 6.035 ~ 6 Na 3 AlF 6
Report as oxides Mineral composition reported as oxides, e.g. Al as Al 2 O 3 Fe 2+ and Fe 3+ often reported together as FeO (i.e., all Fe 2+ ) Conventional order SiO 2, Al 2 O 3, Fe 2 O 3, TiO 2, Cr 2 O 3, FeO MnO, MgO, CaO, BaO, Na 2 O, K 2 O, F, Cl
Between oxide and element wt% Example Mg MgO Molecular weight: Mg = 24.3 MgO = 24.3 + 16 = 40.3 4.55 wt % Mg?? wt% MgO Step 1: Assuming 100 g of sample, mole of Mg: 4.55 / 24.3 = 0.187 Step 2: weight of MnO = 40.3 0.187 = 7.54 g in 100g sample wt% = 7.54%
From oxide wt% to formula Example Again, assuming 100 g of sample wt% mw m.w. Moles Moles Moles oxide cation anion SiO 2 42.7 60.086 42.7/60.086 0.7106 0.7106 1.4213 MgO 57.3 40.312 57.3/40.312 1.4214 1.4214 1.4214 2.8427 Atomic ratio Mg : Si : O = 1.4214 : 0.7106 : 2.8427 = 2 : 1: 4 Mg 2 SiO 4
Step 1 Moles of oxide oxide A B C weight percent Molecular weight A / B (g oxide / 100 g (g oxide / mol oxide) (mol oxide / 100 g mineral) mineral) SiO 2 48.0 60.084 0.80 Al 2 O 3 33.7 101.961 0.33 CaO 17.3 56.079 0.31 Na 2 O 0.63 61.9756 0.00 K 2 O 002 0.02 94.196 001 0.01 Total 99.7
Step 2 Normalize Oxygens C D E 2.908 = Number of oxygens per oxide A / B Oxygen C * D 100 g mineral (mol s / oxide (Oxygens / We know one formula unit of oxide / 100 g 100 g mineral) mineral contains 8 oxygens (for a SiO 2 mineral) 0.80 2 1.598 feldspar) So we renormalize column E to Al 2 O 3 0.33 3 0.992 eight total oxygens, giving us one CaO 0.31 1 0.308 formula unit of mineral. Na 2 O 000 0.00 1 0.000000 Multiply each value by the ratio: K 2 O 0.01 1 0.010 8 oxygens per formula unit Total 2.908 / 2.908 oxygens per 100 g mineral
Step 2 Normalize Oxygens oxide C D E F A / B (mol oxide / 100 g mineral) Oxygens / oxide C * D (Oxygens / 100 g mineral) E * (8/2.908) (Oxygens / formula unit) SiO 2 0.80 2 1.598 4.395 Al 2 O 3 033 0.33 3 0.992 2.728 CaO 0.31 1 0.308 0.849 Na 2 O 0.00 1 0.000 0.001 K 2 O 0.01 1 0.010 0.028 Total 2.908 8.000
Step 3 Calculate Cations oxide Ox / oxide D F G H E * (8/2.908) (Oxygens / formula unit) Cations / oxide (F/D)*G (Cations / formula unit) SiO 2 2 4.395 1 2.198 Al 2 O 3 3 2.728 2 1.818 CaO 1 0.849 1 0.849 Na 2 O 1 0.001 2 0.001 K 2 O 1 0.028028 2 0.056056 Total 8.000 4.922
Step 4 Allocate cations to sites K 0.056 Na 0.001 Ca 0.849 Al 1.818 Si 2.198 O 8 H oxide (F/D)*G (Cations / formula unit) SiO 2 2.198 Al 2 O 3 1.818 CaO 0.849 Na 2 O 0.001001 K 2 O 0.056